linux0.01源码head.s分页机制疑问

目标码神 2013-10-15 09:05:19
下面是head.s的源码
/*
* head.s contains the 32-bit startup code.
*
* NOTE!!! Startup happens at absolute address 0x00000000, which is also where
* the page directory will exist. The startup code will be overwritten by
* the page directory.
*/
.text
.globl idt,gdt,pg_dir
pg_dir:
.globl startup_32
startup_32:
movl $0x10,%eax
mov %ax,%ds
mov %ax,%es
mov %ax,%fs
mov %ax,%gs
lss stack_start,%esp
call setup_idt
call setup_gdt
movl $0x10,%eax # reload all the segment registers
mov %ax,%ds # after changing gdt. CS was already
mov %ax,%es # reloaded in 'setup_gdt'
mov %ax,%fs
mov %ax,%gs
lss stack_start,%esp
xorl %eax,%eax
1: incl %eax # check that A20 really IS enabled
movl %eax,0x000000
cmpl %eax,0x100000
je 1b
movl %cr0,%eax # check math chip
andl $0x80000011,%eax # Save PG,ET,PE
testl $0x10,%eax
jne 1f # ET is set - 387 is present
orl $4,%eax # else set emulate bit
1: movl %eax,%cr0
jmp after_page_tables

/*
* setup_idt
*
* sets up a idt with 256 entries pointing to
* ignore_int, interrupt gates. It then loads
* idt. Everything that wants to install itself
* in the idt-table may do so themselves. Interrupts
* are enabled elsewhere, when we can be relatively
* sure everything is ok. This routine will be over-
* written by the page tables.
*/
setup_idt:
lea ignore_int,%edx
movl $0x00080000,%eax
movw %dx,%ax /* selector = 0x0008 = cs */
movw $0x8E00,%dx /* interrupt gate - dpl=0, present */

lea idt,%edi
mov $256,%ecx
rp_sidt:
movl %eax,(%edi)
movl %edx,4(%edi)
addl $8,%edi
dec %ecx
jne rp_sidt
lidt idt_descr
ret

/*
* setup_gdt
*
* This routines sets up a new gdt and loads it.
* Only two entries are currently built, the same
* ones that were built in init.s. The routine
* is VERY complicated at two whole lines, so this
* rather long comment is certainly needed :-).
* This routine will be overwritten by the page tables.
*/
setup_gdt:
lgdt gdt_descr
ret

.org 0x1000
pg0:

.org 0x2000
pg1:

.org 0x3000
pg2: # This is not used yet, but if you
# want to expand past 8 Mb, you'll have
# to use it.

.org 0x4000
after_page_tables:
pushl $0 # These are the parameters to main :-)
pushl $0
pushl $0
pushl $L6 # return address for main, if it decides to.
pushl $main
jmp setup_paging
L6:
jmp L6 # main should never return here, but
# just in case, we know what happens.

/* This is the default interrupt "handler" :-) */
.align 2
ignore_int:
incb 0xb8000+160 # put something on the screen
movb $2,0xb8000+161 # so that we know something
iret # happened


/*
* Setup_paging
*
* This routine sets up paging by setting the page bit
* in cr0. The page tables are set up, identity-mapping
* the first 8MB. The pager assumes that no illegal
* addresses are produced (ie >4Mb on a 4Mb machine).
*
* NOTE! Although all physical memory should be identity
* mapped by this routine, only the kernel page functions
* use the >1Mb addresses directly. All "normal" functions
* use just the lower 1Mb, or the local data space, which
* will be mapped to some other place - mm keeps track of
* that.
*
* For those with more memory than 8 Mb - tough luck. I've
* not got it, why should you :-) The source is here. Change
* it. (Seriously - it shouldn't be too difficult. Mostly
* change some constants etc. I left it at 8Mb, as my machine
* even cannot be extended past that (ok, but it was cheap :-)
* I've tried to show which constants to change by having
* some kind of marker at them (search for "8Mb"), but I
* won't guarantee that's all :-( )
*/
.align 2
setup_paging:
movl $1024*3,%ecx
xorl %eax,%eax
xorl %edi,%edi /* pg_dir is at 0x000 */
cld;rep;stosl
movl $pg0+7,pg_dir /* set present bit/user r/w */
movl $pg1+7,pg_dir+4 /* --------- " " --------- */
movl $pg1+4092,%edi
movl $0x7ff007,%eax /* 8Mb - 4096 + 7 (r/w user,p) */
std
1: stosl /* fill pages backwards - more efficient :-) */
subl $0x1000,%eax
jge 1b
xorl %eax,%eax /* pg_dir is at 0x0000 */
movl %eax,%cr3 /* cr3 - page directory start */
movl %cr0,%eax
orl $0x80000000,%eax
movl %eax,%cr0 /* set paging (PG) bit */
ret /* this also flushes prefetch-queue */

.align 2
.word 0
idt_descr:
.word 256*8-1 # idt contains 256 entries
.long idt
.align 2
.word 0
gdt_descr:
.word 256*8-1 # so does gdt (not that that's any
.long gdt # magic number, but it works for me :^)

.align 8
idt: .fill 256,8,0 # idt is uninitialized

gdt: .quad 0x0000000000000000 /* NULL descriptor */
.quad 0x00c09a00000007ff /* 8Mb */
.quad 0x00c09200000007ff /* 8Mb */
.quad 0x0000000000000000 /* TEMPORARY - don't use */
.fill 252,8,0 /* space for LDT's and TSS's etc */


请问
setup_paging:
movl $1024*3,%ecx
xorl %eax,%eax
xorl %edi,%edi /* pg_dir is at 0x000 */
cld;rep;stosl
movl $pg0+7,pg_dir /* set present bit/user r/w */
movl $pg1+7,pg_dir+4 /* --------- " " --------- */
movl $pg1+4092,%edi
movl $0x7ff007,%eax /* 8Mb - 4096 + 7 (r/w user,p) */
std
1: stosl /* fill pages backwards - more efficient :-) */
subl $0x1000,%eax
jge 1b
xorl %eax,%eax /* pg_dir is at 0x0000 */
movl %eax,%cr3 /* cr3 - page directory start */
movl %cr0,%eax
orl $0x80000000,%eax
movl %eax,%cr0 /* set paging (PG) bit */
ret /* this also flushes prefetch-queue */

其中 movl $1024*3,%ecx
xorl %eax,%eax
xorl %edi,%edi /* pg_dir is at 0x000 */
cld;rep;stosl
不会把现在正在执行的代码给覆盖了吗?
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神-气 2013-10-16
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引用 1 楼 HappyCodeFly 的回复:
还是我自己回答吧,使用bochs经过调试发现,执行到rep stosw这条指令时,指令内存已经到了物理地址0x402c,而覆盖的地方是1024*3*4字节=12KB,写成16进制也就是0x0000~0x03000处,所以不会有问题的。也许linus是为了重用内存才把这部分内存区域覆盖掉的,毕竟那时候内存还是少的可怜的,所以有这种一点也不浪费的思想也算正常。 给自己加10分!!!!
支持。给我吧。
applelemon_1985 2013-10-16
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目标码神 2013-10-15
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还是我自己回答吧,使用bochs经过调试发现,执行到rep stosw这条指令时,指令内存已经到了物理地址0x402c,而覆盖的地方是1024*3*4字节=12KB,写成16进制也就是0x0000~0x03000处,所以不会有问题的。也许linus是为了重用内存才把这部分内存区域覆盖掉的,毕竟那时候内存还是少的可怜的,所以有这种一点也不浪费的思想也算正常。 给自己加10分!!!!

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